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Let $\vec {a}=2\vec {i}+\vec {j}-\vec {k}$, $\vec {b}=\vec {i}-\vec {j}$ and $\vec {c}=5\vec {i}-\vec {j}+\vec {k}$. Then find a unit vector parallel to $\vec {a}+\vec {b}-\vec {c}$ but in opposite direction.

My Attempt: $$\vec {a}=(2 , 1 , -1)$$ $$\vec {b}=(1 , -1 , 0)$$ $$\vec {c}=(5 , -1 , 1)$$ Now, $$=\vec {a}+\vec {b}+\vec {c}$$ $$=(2 , 1 , -1) + (1 , -1, 0) + (5 , -1, 1)$$ $$(8 , -1 ,0)$$.

Now, what should I do further?

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  • $\begingroup$ i have another vector $$[-2,1,-2]$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 17 '17 at 15:28
  • $\begingroup$ @Dr.SonnhardGraubner, Could you please add.calculations? $\endgroup$ – pi-π Apr 17 '17 at 15:31
  • $\begingroup$ @Ramanujan it appears you made a calculation error when finding $\vec{a}+\vec{b}-\vec{c}$ $\endgroup$ – Tom Carter Apr 17 '17 at 15:46
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You computed $\vec{a}+\vec{b}+\vec{c}$ instead of $\vec{a}+\vec{b}-\vec{c}$. Let's call the result of the latter calculation $\vec{V}$. Then $\frac{\vec{V}}{\vec{V}\cdot\vec{V}}$ is a unit vector in the same direction as $\vec{V}$. As we've been asked for the opposite direction, we need $-\frac{\vec{V}}{\vec{V}\cdot\vec{V}}$. Equivalently, you could compute $\frac{\vec{W}}{\vec{W}\cdot\vec{W}}$ with $\vec{W}:=-\vec{V}=\vec{c}-\vec{a}-\vec{b}$.

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the searched vector is given by $$\vec{a}+\vec{b}-\vec{c}=(2+1-5;1-1+1;-1-1)$$

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Then find $\dfrac{-(\vec{a}+\vec{b}-\vec{c})}{||\vec{a}+\vec{b}-\vec{c}||}$ that's the unit vector

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